3.170 \(\int (a \sin (e+f x))^m (b \tan (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=79 \[ \frac {2 \cos ^2(e+f x)^{5/4} (b \tan (e+f x))^{5/2} (a \sin (e+f x))^m \, _2F_1\left (\frac {5}{4},\frac {1}{4} (2 m+5);\frac {1}{4} (2 m+9);\sin ^2(e+f x)\right )}{b f (2 m+5)} \]

[Out]

2*(cos(f*x+e)^2)^(5/4)*hypergeom([5/4, 5/4+1/2*m],[9/4+1/2*m],sin(f*x+e)^2)*(a*sin(f*x+e))^m*(b*tan(f*x+e))^(5
/2)/b/f/(5+2*m)

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Rubi [A]  time = 0.12, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2602, 2577} \[ \frac {2 \cos ^2(e+f x)^{5/4} (b \tan (e+f x))^{5/2} (a \sin (e+f x))^m \, _2F_1\left (\frac {5}{4},\frac {1}{4} (2 m+5);\frac {1}{4} (2 m+9);\sin ^2(e+f x)\right )}{b f (2 m+5)} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sin[e + f*x])^m*(b*Tan[e + f*x])^(3/2),x]

[Out]

(2*(Cos[e + f*x]^2)^(5/4)*Hypergeometric2F1[5/4, (5 + 2*m)/4, (9 + 2*m)/4, Sin[e + f*x]^2]*(a*Sin[e + f*x])^m*
(b*Tan[e + f*x])^(5/2))/(b*f*(5 + 2*m))

Rule 2577

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b^(2*IntPart
[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*(a*Sin[e + f*x])^(m + 1)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rule 2602

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Cos[e + f
*x]^(n + 1)*(b*Tan[e + f*x])^(n + 1))/(b*(a*Sin[e + f*x])^(n + 1)), Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^
n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int (a \sin (e+f x))^m (b \tan (e+f x))^{3/2} \, dx &=\frac {\left (a \cos ^{\frac {5}{2}}(e+f x) (b \tan (e+f x))^{5/2}\right ) \int \frac {(a \sin (e+f x))^{\frac {3}{2}+m}}{\cos ^{\frac {3}{2}}(e+f x)} \, dx}{b (a \sin (e+f x))^{5/2}}\\ &=\frac {2 \cos ^2(e+f x)^{5/4} \, _2F_1\left (\frac {5}{4},\frac {1}{4} (5+2 m);\frac {1}{4} (9+2 m);\sin ^2(e+f x)\right ) (a \sin (e+f x))^m (b \tan (e+f x))^{5/2}}{b f (5+2 m)}\\ \end {align*}

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Mathematica [A]  time = 8.14, size = 87, normalized size = 1.10 \[ \frac {2 (b \tan (e+f x))^{5/2} \sec ^2(e+f x)^{m/2} (a \sin (e+f x))^m \, _2F_1\left (\frac {m+2}{2},\frac {1}{4} (2 m+5);\frac {1}{4} (2 m+9);-\tan ^2(e+f x)\right )}{b f (2 m+5)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sin[e + f*x])^m*(b*Tan[e + f*x])^(3/2),x]

[Out]

(2*Hypergeometric2F1[(2 + m)/2, (5 + 2*m)/4, (9 + 2*m)/4, -Tan[e + f*x]^2]*(Sec[e + f*x]^2)^(m/2)*(a*Sin[e + f
*x])^m*(b*Tan[e + f*x])^(5/2))/(b*f*(5 + 2*m))

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \tan \left (f x + e\right )} \left (a \sin \left (f x + e\right )\right )^{m} b \tan \left (f x + e\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))^m*(b*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*tan(f*x + e))*(a*sin(f*x + e))^m*b*tan(f*x + e), x)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))^m*(b*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 0.44, size = 0, normalized size = 0.00 \[ \int \left (a \sin \left (f x +e \right )\right )^{m} \left (b \tan \left (f x +e \right )\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sin(f*x+e))^m*(b*tan(f*x+e))^(3/2),x)

[Out]

int((a*sin(f*x+e))^m*(b*tan(f*x+e))^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}} \left (a \sin \left (f x + e\right )\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))^m*(b*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e))^(3/2)*(a*sin(f*x + e))^m, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sin(e + f*x))^m*(b*tan(e + f*x))^(3/2),x)

[Out]

int((a*sin(e + f*x))^m*(b*tan(e + f*x))^(3/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))**m*(b*tan(f*x+e))**(3/2),x)

[Out]

Timed out

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